$\text { Find } \frac { d y } { d x }$ $\begin{array} { l }
x = \sqrt [ 12 ] { t } \
y = 6 - t
\end{array}$
A)$\frac { d y } { d x } = - 12 t ^ { \frac { 13 } { 12 } }$
B)$\frac { d y } { d x } = - 12 t ^ { \frac { 11 } { 12 } }$
C)$\frac { d y } { d x } = - 6 t ^ { \frac { 11 } { 12 } }$
D)$\frac { d y } { d x } = 6 t ^ { \frac { 13 } { 12 } }$
E)$\frac { d y } { d x } = 12 t ^ { \frac { 11 } { 12 } }$

Question

$\text { Find } \frac { d y } { d x }$ $\begin{array} { l } 
x = \sqrt [ 12 ] { t } \
y = 6 - t
\end{array}$
A)$\frac { d y } { d x } = - 12 t ^ { \frac { 13 } { 12 } }$
B)$\frac { d y } { d x } = - 12 t ^ { \frac { 11 } { 12 } }$
C)$\frac { d y } { d x } = - 6 t ^ { \frac { 11 } { 12 } }$
D)$\frac { d y } { d x } = 6 t ^ { \frac { 13 } { 12 } }$
E)$\frac { d y } { d x } = 12 t ^ { \frac { 11 } { 12 } }$

Quizplus · Accepted Answer

To find $\frac{dy}{dx}$, we use the chain rule, which in this context is $\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}$. First, differentiate $y$ with respect to $t$, which gives $\frac{dy}{dt} = -1$, since $y = 6 - t$. Then, differentiate $x$ with respect to $t$, where $x = t^{\frac{1}{12}}$, giving $\frac{dx}{dt} = \frac{1}{12}t^{-\frac{11}{12}}$. To find $\frac{dt}{dx}$, we take the reciprocal of $\frac{dx}{dt}$, which is $12t^{\frac{11}{12}}$. Thus, $\frac{dy}{dx} = -1 \cdot 12t^{\frac{11}{12}} = -12t^{\frac{11}{12}}$.

Find dydx\text { Find } \frac { d y } { d x } Find dxdy​

$\text { Find } \frac { d y } { d x }$